Math1013 Calculus IB, Fall 2017
Final Exam Answers
Part I: Multiple Choice Questions.
Question
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18
White
C
C
D
B
C
E
A
E
C
B
B
A
D
E
E
A
D
A
Part II: Long Questions.
19. Let r be the radius, and h the height, of the right circular cylinder. Then r 2 + ( h2 )2 = 42 , and the
surface area of the cylinder is
p
A = 2πr 2 + 2πr · 2 16 − r 2
where 0 ≤ r ≤ 4.
√
h
p
dA
2r 2 i
16 + r 16 − r 2 − 2r 2
2
√
= 2π 2r + 2 16 − r − √
=0
= 4π ·
dr
16 − r 2
16 − r 2
√
if and only if 2r 2 − 16 = r 16 − r 2 .
Squaring both sides, we have
5r 4 − 80r 2 − 256 = 0
√
√
80 ± 802 − 20 · 162
8 5
2
r =
=8±
10
5
√
√
Note that r 2 = 8 − 8 5 5 does not satisfy the equation 2r 2 − 16 = r 16 − r 2 . Hence the only critical
point of A(r) in the interval [0, 4] is
s
√
8 5
r = 8+
5
Noting that

q
√
 > 0 if 0 < r < 8 + 8 5
5
q
A′ (r)
√
 < 0 if 8 + 8 5 < r < 4
5
the largest surface area is reached at
r=
by the first derivative test.
s
√
8 5
8+
5
s
√
p
8
5
h = 2 16 − r 2 = 2 8 −
5
20. (a) Differentiating both sides of the equation,
x3
dy
dx
dy
dx
+ 3yx2
+ 2y
= 4x
dt
dt
dt
dt
dy
dx
= (4x − 3yx2 )
dt
dt
dy
4−3
=
·6=2
dt (1,1) 1 + 2
(x3 + 2y)
(b) Rising since
dy
dt
= 2 > 0.
(1,1)
1
21. (a) By checking the sign of f ′ (x), we have:
Interval of increase: (−∞, −2) or (−2, 0]
Interval of decrease: [0, 2) or (2, ∞)
(b) By checking the sign of f ′′ (x), we have:
Concave up interval: (−∞, −2) or (2, ∞)
(c)
y
4
3
2
1
x
−4
−3
−2
−1
1
2
3
4
−1
−2
−3
−4
22. (a) Let u = a − x such that du = −dx. Then
Z a
Z 0
Z
f (x)dx =
−f (a − u)du =
0
0
a
a
f (a − u)du =
Z
a
0
f (a − x)dx
na
(b) Using the subdivision points 0, na , 2a
n , . . . , n of the interval [0, a], and the right-endpoint of
each subinterval, and left-endpoint respectively, we have
Z a
ah
a
2a
3a
(n − 1)a
na i
f (a − x)dx = lim
f (a − ) + f (a − ) + f (a − ) + · · · + f (a −
) + f (a − )
n→∞ n
n
n
n
n
n
0
and respectively
Z a
ah
a
(n − 2)a
(n − 1)a i
f (x)dx = lim
f (0) + f ( ) + · · · + f (
) + f(
)
n→∞ n
n
n
n
0
The two limits must be the same since by reordering the terms,
f (a −
a
2a
3a
(n − 1)a
na
) + f (a − ) + f (a − ) + · · · + f (a −
) + f (a −
)
n
n
n
n
n
(n − 1)a
(n − 2)a
a
) + f(
) + · · · + f ( ) + f (0)
n
n
n
a
(n − 2)a
(n − 1)a
= f (0) + f ( ) + · · · + f (
) + f(
)
n
n
n
= f(
(c) Using the result in (a),
Z
0
=
i.e.,
Z
π
2
0
π
2
3 cos x − sin x
dx =
cos x + sin x
3 sin x − cos x
dx =
sin x + cos x
Z
π
2
0
Z
π
2
0
Z
0
π
2
3 cos( π2 − x) − sin( π2 − x)
dx
cos( π2 − x) + sin( π2 − x)
sin x − 3 cos x
dx +
sin x + cos x
3 cos x − sin x
dx =
cos x + sin x
Z
π
2
0
Z
π
2
0
1dx =
π
2
2 sin x + 2 cos x
dx
sin x + cos x